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 Incoming links: Dirac delta scaling

Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 2 / 7A / a / i / Solution Created 2026-09-24 Updated 2026-10-03

For every test function

φ

and nonzero

p

, substitution

u = pt

with the orientation handled by

∣ p ∣

gives

\int φ (t) δ (pt) d t = \frac{1}{∣ p ∣} φ (0) .

Hence the Dirac delta scaling identity is $δ (pt) = δ (t) /∣ p ∣$ .

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